If runs of R is 70 and sum of runs of R and S is 120,

Solution

R + S = 120

70 + S = 120

S = 120 – 70 = 50.


Possible Runs of U will between 50 and 70.

The runs of Q is more than the runs of T. The number of persons more than Q is one less than the number of persons less than T.

Q > ___ > _____ > ___ T > ___ ……… (I)

____ > Q > ___ > T > ___ > ___ ……… (II)

The runs of R is more than U, whose runs is more than S. At least two persons have less runs than S.

Q > R > U > S > T > P ……… (I)

____ > Q > ___ > T > ___ > ___ ……… (II)


Case 2 will get discarded.


Therefore, the final arrangement is as follow:

Q(highest) > R > U > S > T > P(lowest)